Question

the position of an object changes with time as X = 2t- 5t^2 + t^3 . the initial velocity of object is​

Answer 1

Given:

The position of an object changes with time as X = 2t- 5t^2 + t^3

To find:

Initial velocity of object

Calculation:

The Velocity function can be calculated by first-order differentiation of the displacement function with respect to time.

$\therefore \: v = \dfrac{dx}{dt}$

$= > \: v = \dfrac{d(2t - 5 {t}^{2} + {t}^{3} )}{dt}$

$= > \: v = \dfrac{d(2t )}{dt} - 5 \dfrac{d( {t}^{2}) }{dt} + \dfrac{d( {t}^{3}) }{dt}$

$= > \: v = 2 - (5 \times 2) + 3 {t}^{2}$

$= > \: v = 2 - 10 + 3 {t}^{2}$

$= > \: v = 3 {t}^{2} - 8$

So, initial velocity can be calculated by putting t = 0 sec.

$= > \: v_{0}= 3 {(0)}^{2} - 8$

$= > \: v_{0}= 0 - 8$

$= > \: v_{0}= - 8 \: m {s}^{ - 1}$

So, final answer is:

$\boxed{ \bf{\: v_{0}= - 8 \: m {s}^{ - 1} }}$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

If a particle travels a distance x at a time t then it's velocity at any time t is given by

$\displaystyle \sf{ v = \frac{dx}{dt} \: \: }$

GIVEN

The position of an object changes with time as

$\sf{x = 2t - 5 {t}^{2} + {t}^{3} \: \: }$

TO DETERMINE

The initial velocity of the object

CALCULATION

Here

$\sf{x = 2t - 5 {t}^{2} + {t}^{3} \: \: }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dx}{dt} =\frac{d}{dt}(2t) - \frac{d}{dt}(5 {t}^{2} ) + \: \frac{d}{dt} ( {t}^{3} )\: \: }$

$\implies \: \displaystyle \sf{ v = \frac{dx}{dt} =2\frac{d}{dt}(t) - 5 \frac{d}{dt}( {t}^{2} ) + \: \frac{d}{dt} ( {t}^{3} )\: \: }$

$\implies \: \displaystyle \sf{ v = \frac{dx}{dt} =2 - (5 \times 2t) + \: 3 {t}^{2} \: \: }$

$\implies \: \displaystyle \sf{ v = \frac{dx}{dt} =2 - 10t + \: 3 {t}^{2} \: \: }$

Hence the initial velocity is obtained by putting t = 0

$\displaystyle \: \sf{ Initial \: Velocity = \bigg [ \frac{dx}{dt} \bigg] _{(t=0)} =2 - 0 + 0 = 2\: \: }$

RESULT

The Initial Velocity is 2 unit

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